What is the solution for 2x^{''}-x^'+3t=5 ?

The solution for 2x^{''}-x^'+3t=5 is x=c_{1}e^{t/2}+c_{2}+(3t^2)/2+t

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2x^{''}-x^'+3t=5

2 x^{^{''}} - x^{^{'}} + 3 t = 5

x = c_{1} e^{\frac{t}{2}} + c_{2} + \frac{3 t ^{2}}{2} + t

Solution steps

2 x^{''} (t) - x^{'} (t) + 3 t = 5

Solve linear ODE:

x = c_{1} e^{\frac{t}{2}} + c_{2} + \frac{3 t ^{2}}{2} + t

x = c_{1} e^{\frac{t}{2}} + c_{2} + \frac{3 t ^{2}}{2} + t

y^{^{''}} - 2 y^{^{'}} + 82 y = e x cos (9 x)

y^{^{''}} - 9 y^{^{'}} + 6 y = x e^{x}

y^{^{''}} - 2 y^{^{'}} + 2 y = 5 x^{2} - 9 x + 5

y^{^{''}} + 5 y = - 5 - 6 x + 12 x^{2} - 5 x^{3} + 5 x^{4}

y^{^{''}} + 5 y^{^{'}} - 6 y = 36 x + 98 x e^{x}

What is the solution for 2x^{''}-x^'+3t=5 ?
The solution for 2x^{''}-x^'+3t=5 is x=c_{1}e^{t/2}+c_{2}+(3t^2)/2+t